Proposition 27

Table of Contents

PROPOSITION 27 PROBLEM

For the point A (Fig. 22) is to be moved forwards through the distance AP by a uniform force, and it is required to find the time in which the distance AP is completed.

SOLUTION

As before, let the force acting be g, the distance AP = x and the height corresponding to the speed at P should be v; on this account …

Therefore the speed itself at P is equal to .. v = The time therefore be found, in which the element Pp = dx is traversed, varies as gx Adx .

Let the time in which … the distance AP is completed be t and put , …

it is necessary to perform a single experiment to determine the value of the letter [p. 87] m, so that the time in the given measurement can be found, reckoned in seconds. Indeed from that equation the time t will be produced by integration and we have ,2 g … xmt = to which there is no need to add on a constant quantity, since for the position x = 0 the time t vanishes, as it should.

Therefore on determining m by experiment we have the time to fall given by g Axmt 2= … sec.

Moreover, so that the measurement of the time by this means shall be absolute, it is necessary that x likewise is shown to be measured in agreement with this constant [i. e. in some standard unit of length]:

Therefore we will always determine the interval x in scruples, i. e. thousandth parts of Rhenish feet ; for indeed the fraction g A is expressed in absolute numbers, in order that it will not be necessary to have a certain measurement for that. Therefore with the letter m defined, that we will make soon, the complete solution to the problem will be had. Q. E. I.

Corollary 1

If g designates the force of gravity, then 1=g A (205) ; on account of this the time in which the body will fall to the earth from the height of x scruples of Rhenish feet, will be xm2 seconds.

Corollary 2

Moreover from experiments it has been ascertained that in a time of one second a body falls through a height of 15625 scruples of Rhenish feet [4.904 metres, which is in good agreement with the modern accepted value, although we do not know where the measurements were made for the full significant figures]. When on account of this, if the height is put as x = 15625, a time of one second will be produced : t = 1.

But since xmt 2= , then 1562521 m= , i. e. = 250m. Hence the letter m is found to be 250 1=m .

Corollary 3

Since indeed the letter m retains the same value in all cases, in the case of the problem, the time will be g .. Axt 125 1= sec.

Therefore with the distance x expressed in scruples of Rhenish feet, the time as the number of seconds is given by g … 1 for this space to be traversed [by falling from rest].

Corollary 4

And in all straight forward cases this value of m found can be applied. Indeed let the element of distance to be described be ds, with the height v corresponding to the speed in which this is traversed, the element of time is .and …

From this equation, if v and s are expressed in scruples of Rhenish feet and we put 250 1=m , the time will be produced in seconds : … 1 sec.

Scholium 1

Therefore from this, since we specify the speeds by the square roots of the corresponding heights, which we take again for convenience in the following, as we always find the measure of the absolute time.

For truly the experiment we have used from which the height is found that a weight falls in a time of one second, as Huygens found by experiments with pendulums to be 15 Paris feet, 1 digit, 18 12 lines, i. e. in decimal fractions 15.0796 Parisian feet. Moreover the ratio of Rhenish feet to Parisian feet we use is 1000 to 1035, from which the height to fall in one second comes to be 15.625 Rhenish feet or 15,625 [p. 89] scruples of the same feet; and this is the measure we prefer to have rather than the Parisian one, since this number is a square and in this way we avoid frequent root extractions.

Besides, the number 250 that is found is easily remembered, by which ∫ v ds (with s and v expressed in scruples of Rhenish feet) must be divided in order to find the time in seconds.

Corollary 5

Since A g denotes the strength of the acceleration of the force (213), the times will be, in which any distances are traversed under uniform forces, in the square root ratio composed from the distances and inversely as the strengths of the accelerations.

Corollary 6

With the speed c put in place, that the point A acquires with a force g acting over the height x; c varies as A gx (193). Hence ct will be as x, since t is as g Ax .

Consequently, the distances travelled through are in the ratio composed from the times in which they are described, and the speed that they gain in the descent, for whatever the forces acting in a uniform manner.

Corollary 7

And the distances that are traversed in equal intervals of time are as the strengths of the accelerations of the forces acting. [p. 90]

Corollary 8

Therefore the distances through which bodies fall in equal intervals of time on the surfaces of the Sun, Jupiter, Saturn, the Moon, and the Earth, are amongst themselves as 24390, 2036, 1280, 333, 1000. (214).

Corollary 9

For the hypothesis of the same acceleration under the same force, the times in which any distances are traversed , are as the speeds acquired, and so for the times as with the speeds, are in the ratio of the square roots of the distances described.

Scholium 2

Here we have always put the descending bodies to start the descent from rest or their initial speed to be zero.

In the following indeed we will investigate these motions, which arise when the bodies are themselves in motion at the start and now they have some speed.

With these both the times and the distances ought to be understood, which in the beginning have their own point where the speed vanishes, and all the equations found are thus comparable, as with the vanishing of c or v likewise x and t vanish.